SECTION 5: LAPLACE TRANSFORMS
ESE 330 – Modeling & Analysis of Dynamic Systems 2 Introduction – Transforms
This section of notes contains an introduction to Laplace transforms. This should mostly be a review of material covered in your differential equations course.
K. Webb ESE 330 Transforms 3
What is a transform? A mapping of a mathematical function from one domain to another A change in perspective not a change of the function Why use transforms? Some mathematical problems are difficult to solve in their natural domain Transform to and solve in a new domain, where the problem is simplified Transform back to the original domain Trade off the extra effort of transforming/inverse- transforming for simplification of the solution procedure
K. Webb ESE 330 Transform Example – Slide Rules 4
Slide rules make use of a logarithmic transform
Multiplication/division of large numbers is difficult Transform the numbers to the logarithmic domain Add/subtract (easy) in the log domain to multiply/divide (difficult) in the linear domain Apply the inverse transform to get back to the original domain Extra effort is required, but the problem is simplified
K. Webb ESE 330 5 Laplace Transforms
K. Webb ESE 330 Laplace Transforms 6
An integral transform mapping functions from the time domain to the Laplace domain or s-domain
ℒ Time-domain functions𝑔𝑔 𝑡𝑡 are↔ functions𝐺𝐺 𝑠𝑠 of time, 𝑡𝑡 Laplace-domain functions𝑔𝑔 are𝑡𝑡 functions of 𝒔𝒔 is a complex variable 𝐺𝐺 𝑠𝑠 𝑠𝑠 = + K. Webb 𝑠𝑠 𝜎𝜎 𝑗𝑗𝑗𝑗 ESE 330 Laplace Transforms – Motivation 7
We’ll use Laplace transforms to solve differential equations
Differential equations in the time domain difficult to solve Apply the Laplace transform Transform to the s-domain Differential equations become algebraic equations easy to solve Transform the s-domain solution back to the time domain
Transforming back and forth requires extra effort, but the solution is greatly simplified
K. Webb ESE 330 Laplace Transform 8
Laplace Transform:
= = (1) ∞ −𝑠𝑠𝑠𝑠 Unilateralℒ or𝑔𝑔 one𝑡𝑡 -sided𝐺𝐺 𝑠𝑠transform∫0 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 Lower limit of integration is = 0 Assumed that the time domain function is zero for all negative time, i.e. 𝑡𝑡 = 0, < 0
𝑔𝑔 𝑡𝑡 𝑡𝑡
K. Webb ESE 330 9 Laplace Transform Properties
In the following section of notes, we’ll derive a few important properties of the Laplace transform.
K. Webb ESE 330 Laplace Transform – Linearity 10
Say we have two time-domain functions: and
Applying the𝑔𝑔 1transform𝑡𝑡 𝑔𝑔 2definition,𝑡𝑡 (1) + = + ∞ −𝑠𝑠𝑠𝑠 ℒ 𝛼𝛼𝑔𝑔1 𝑡𝑡 𝛽𝛽𝑔𝑔2 𝑡𝑡 � 𝛼𝛼𝑔𝑔1 𝑡𝑡 𝛽𝛽𝑔𝑔2 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 = 0 + ∞ ∞ −𝑠𝑠𝑠𝑠 −𝑠𝑠𝑠𝑠 � 𝛼𝛼𝑔𝑔1 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 � 𝛽𝛽𝑔𝑔2 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 = 0 + 0 ∞ ∞ −𝑠𝑠𝑠𝑠 −𝑠𝑠𝑠𝑠 1 2 = 𝛼𝛼 �0 𝑔𝑔 𝑡𝑡 𝑒𝑒 + 𝑑𝑑𝑑𝑑 𝛽𝛽 �0 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑
𝛼𝛼 � ℒ 𝑔𝑔1 𝑡𝑡 𝛽𝛽 � ℒ 𝑔𝑔2 𝑡𝑡 + = + (2)
The ℒLaplace𝛼𝛼𝑔𝑔1 transform𝑡𝑡 𝛽𝛽𝑔𝑔 2is a𝑡𝑡 linear 𝛼𝛼operation𝐺𝐺1 𝑠𝑠 𝛽𝛽𝐺𝐺2 𝑠𝑠 K. Webb ESE 330 Laplace Transform of a Derivative 11 Of particular interest, given that we want to use Laplace transform to solve differential equations
= ∞ −𝑠𝑠𝑠𝑠 Use integrationℒ 𝑔𝑔bẏ 𝑡𝑡 parts�to0 evaluate𝑔𝑔̇ 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 = Let = ∫ 𝑢𝑢𝑢𝑢𝑢𝑢 and𝑢𝑢𝑢𝑢 − ∫ 𝑣𝑣𝑣𝑣𝑣𝑣= then −𝑠𝑠𝑠𝑠 𝑢𝑢 𝑒𝑒 𝑑𝑑𝑑𝑑 𝑔𝑔̇ 𝑡𝑡 𝑑𝑑𝑑𝑑 = and = −𝑠𝑠𝑠𝑠 K. Webb 𝑑𝑑𝑑𝑑 −𝑠𝑠𝑒𝑒 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑔𝑔 𝑡𝑡 ESE 330 Laplace Transform of a Derivative 12
= ∞ ∞ −𝑠𝑠𝑠𝑠 −𝑠𝑠𝑠𝑠 ℒ 𝑔𝑔̇ 𝑡𝑡 𝑒𝑒 𝑔𝑔 𝑡𝑡 �0 − �0 𝑔𝑔 𝑡𝑡 −𝑠𝑠𝑒𝑒 𝑑𝑑𝑑𝑑 = 0 0 + = 0 + ∞ −𝑠𝑠𝑠𝑠 − 𝑔𝑔 𝑠𝑠 �0 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 −𝑔𝑔 𝑠𝑠� 𝑔𝑔 𝑡𝑡 The Laplace transform of the derivative of a function is the Laplace transform of that function multiplied by minus the initial value of that function 𝑠𝑠 = (0) (3)
ℒ 𝑔𝑔̇ 𝑡𝑡 𝑠𝑠𝑠𝑠 𝑠𝑠 − 𝑔𝑔 K. Webb ESE 330 Higher-Order Derivatives 13
The Laplace transform of a second derivative is
= 0 0 (4) 2 ℒ 𝑔𝑔̈ 𝑡𝑡 𝑠𝑠 𝐺𝐺 𝑠𝑠 − 𝑠𝑠𝑠𝑠 − 𝑔𝑔̇ In general, the Laplace transform of the derivative of a function is given by 𝒕𝒕𝒕𝒕 𝒏𝒏
= 0 0 0 (5) 𝑛𝑛 𝑛𝑛 𝑛𝑛−1 𝑛𝑛−2 𝑛𝑛−1 ℒ 𝑔𝑔 𝑠𝑠 𝐺𝐺 𝑠𝑠 − 𝑠𝑠 𝑔𝑔 − 𝑠𝑠 𝑔𝑔̇ − ⋯ − 𝑔𝑔
K. Webb ESE 330 Laplace Transform of an Integral 14
The Laplace Transform of a definite integral of a function is given by
= (6) 𝑡𝑡 1 ℒ ∫0 𝑔𝑔 𝜏𝜏 𝑑𝑑𝑑𝑑 𝑠𝑠 𝐺𝐺 𝑠𝑠 Differentiation in the time domain corresponds to multiplication by in the Laplace domain Integration in the time domain corresponds to division by in the𝒔𝒔 Laplace domain
K. Webb 𝒔𝒔 ESE 330 Laplace Transforms of Common 15 Functions Next, we’ll derive the Laplace transform of some common mathematical functions
K. Webb ESE 330 Unit Step Function 16 A useful and common way of characterizing a linear system is with its step response The system’s response (output) to a unit step input The unit step function or Heaviside step function: 0, < 0 1 = 1, 0 𝑡𝑡 𝑡𝑡 � 𝑡𝑡 ≥
K. Webb ESE 330 Unit Step Function – Laplace Transform 17 Using the definition of the Laplace transform
1 = 1 = ∞ ∞ −𝑠𝑠𝑠𝑠 −𝑠𝑠𝑠𝑠 ℒ 𝑡𝑡 � 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 � 𝑒𝑒 𝑑𝑑𝑑𝑑 1 0 1 0 1 = = 0 = ∞ −𝑠𝑠𝑠𝑠 − 𝑒𝑒 �0 − − 𝑠𝑠 𝑠𝑠 𝑠𝑠 The Laplace transform of the unit step
1 = (7) 1 Note that the unilateralℒ Laplace𝑡𝑡 𝑠𝑠transform assumes that the signal being transformed is zero for < 0 Equivalent to multiplying any signal by a unit step 𝑡𝑡 K. Webb ESE 330 Unit Ramp Function 18
The unit ramp function is a useful input signal for evaluating how well a system tracks a constantly- increasing input The unit ramp function:
0, < 0 = , 0 𝑡𝑡 𝑔𝑔 𝑡𝑡 � 𝑡𝑡 𝑡𝑡 ≥
K. Webb ESE 330 Unit Ramp Function – Laplace Transform
19
Could easily evaluate the transform integral Requires integration by parts Alternatively, recognize the relationship between the unit ramp and the unit step Unit ramp is the integral of the unit step Apply the integration property, (6) 1 1 = 1 = 𝑡𝑡 ℒ 𝑡𝑡 ℒ � 𝜏𝜏 𝑑𝑑𝑑𝑑 � 0 𝑠𝑠 𝑠𝑠 = (8) 1 2 K. Webb ℒ 𝑡𝑡 𝑠𝑠 ESE 330 Exponential – Laplace Transform 20 = −𝑎𝑎𝑎𝑎 Exponentials are common𝑔𝑔 𝑡𝑡 components𝑒𝑒 of the responses of dynamic systems
= = ( ) ∞ ∞ −𝑎𝑎𝑎𝑎 −𝑎𝑎𝑎𝑎 −𝑠𝑠𝑠𝑠 − 𝑠𝑠+𝑎𝑎 𝑡𝑡 ℒ 𝑒𝑒 � 𝑒𝑒 𝑒𝑒 𝑑𝑑𝑑𝑑 � 𝑒𝑒 𝑑𝑑𝑑𝑑 0 01 = = 0 − +𝑠𝑠+𝑎𝑎 𝑡𝑡 ∞ + 𝑒𝑒 − �0 − − 𝑠𝑠 𝑎𝑎 𝑠𝑠 𝑎𝑎 = (9) −𝑎𝑎𝑎𝑎 1
K. Webb ℒ 𝑒𝑒 𝑠𝑠+𝑎𝑎 ESE 330 Sinusoidal functions 21
Another class of commonly occurring signals, when dealing with dynamic systems, is sinusoidal signals – both sin and cos
𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔= sin
Recall Euler’s formula𝑔𝑔 𝑡𝑡 𝜔𝜔𝜔𝜔 = cos + sin 𝑗𝑗𝑗𝑗𝑗𝑗 From which it𝑒𝑒 follows that𝜔𝜔𝜔𝜔 𝑗𝑗 𝜔𝜔𝜔𝜔
sin = 𝑗𝑗𝑗𝑗𝑗𝑗 −𝑗𝑗𝑗𝑗𝑗𝑗 𝑒𝑒 − 𝑒𝑒 K. Webb 𝜔𝜔𝜔𝜔 ESE 330 2𝑗𝑗 Sinusoidal functions 22 1 sin = ∞ 𝑗𝑗𝑗𝑗𝑡𝑡 −𝑗𝑗𝑗𝑗𝑡𝑡 −𝑠𝑠𝑠𝑠 ℒ 𝜔𝜔𝜔𝜔1 �0 𝑒𝑒 − 𝑒𝑒 𝑒𝑒 𝑑𝑑𝑑𝑑 = 2𝑗𝑗 ∞ − 𝑠𝑠−𝑗𝑗𝑗𝑗 𝑡𝑡 − 𝑠𝑠+𝑗𝑗𝑗𝑗 𝑡𝑡 1 �0 𝑒𝑒 − 𝑒𝑒 1 𝑑𝑑𝑑𝑑 = 2𝑗𝑗 ∞ ∞ − 𝑠𝑠−𝑗𝑗𝑗𝑗 𝑡𝑡 − 𝑠𝑠+𝑗𝑗𝑗𝑗 𝑡𝑡 � 𝑒𝑒 𝑑𝑑𝑑𝑑 − � 𝑒𝑒 𝑑𝑑𝑑𝑑 2𝑗𝑗1 0 2𝑗𝑗1 0 = − 𝑠𝑠−𝑗𝑗𝑗𝑗 𝑡𝑡 ∞ − 𝑠𝑠++𝑗𝑗𝑗𝑗 𝑡𝑡 ∞ 𝑒𝑒 𝑒𝑒 1 1 �0 −1 1 �01 = 2𝑗𝑗 0−+𝑠𝑠 − 𝑗𝑗𝑗𝑗 2𝑗𝑗0 +− 𝑠𝑠 𝑗𝑗𝑗𝑗 = + + 2𝑗𝑗𝑗𝑗 − 2 2 2𝑗𝑗 𝑠𝑠 − 𝑗𝑗𝑗𝑗 2𝑗𝑗 𝑠𝑠 𝑗𝑗𝑗𝑗 2𝑗𝑗 𝑠𝑠 𝜔𝜔 sin = (10) 𝜔𝜔 2 2 K. Webb ℒ 𝜔𝜔𝜔𝜔 𝑠𝑠 +𝜔𝜔 ESE 330 Sinusoidal functions 23
It can similarly be shown that
cos = (11) 𝑠𝑠 2 2 ℒ 𝜔𝜔𝜔𝜔 𝑠𝑠 +𝜔𝜔
Note that for neither sin( ) nor cos is the function equal to zero for < 0 as the Laplace transform assumes 𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 𝑡𝑡 Really, what we’ve derived is 1 sin and 1 cos
K. Webb ℒ 𝑡𝑡 � 𝜔𝜔𝜔𝜔 ℒ 𝑡𝑡 � 𝜔𝜔𝜔𝜔 ESE 330 24 More Properties and Theorems
K. Webb ESE 330 Multiplication by an Exponential, 25 −𝑎𝑎𝑎𝑎 We’ve seen that = 𝑒𝑒 −𝑎𝑎𝑎𝑎 1 What if another functionℒ 𝑒𝑒 is𝑠𝑠 +multiplied𝑎𝑎 by the decaying exponential term?
= = ∞ ∞ −𝑎𝑎𝑎𝑎 −𝑎𝑎𝑎𝑎 −𝑠𝑠𝑠𝑠 − 𝑠𝑠+𝑎𝑎 𝑡𝑡 ℒ 𝑔𝑔 𝑡𝑡 𝑒𝑒 � 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝑒𝑒 𝑑𝑑𝑑𝑑 � 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝑑𝑑𝑑𝑑 This is just the Laplace0 transform of0 with replaced by + 𝑔𝑔 𝑡𝑡 𝑠𝑠
𝑠𝑠 𝑎𝑎 = + (12) −𝑎𝑎𝑎𝑎 K. Webb ℒ 𝑔𝑔 𝑡𝑡 𝑒𝑒 𝐺𝐺 𝑠𝑠 𝑎𝑎 ESE 330 Decaying Sinusoids 26
The Laplace transform of a sinusoid is
sin = + 𝜔𝜔 ℒ 𝜔𝜔𝜔𝜔 2 2 And, multiplication by an decaying𝑠𝑠 𝜔𝜔 exponential, , results in a substitution of + for , so −𝑎𝑎𝑎𝑎 𝑒𝑒 sin = 𝑠𝑠 𝑎𝑎 𝑠𝑠 + + −𝑎𝑎𝑎𝑎 𝜔𝜔 and ℒ 𝑒𝑒 𝜔𝜔𝜔𝜔 2 2 𝑠𝑠 𝑎𝑎+ 𝜔𝜔 cos = + + −𝑎𝑎𝑎𝑎 𝑠𝑠 𝑎𝑎 2 2 K. Webb ℒ 𝑒𝑒 𝜔𝜔𝜔𝜔 ESE 330 𝑠𝑠 𝑎𝑎 𝜔𝜔 Time Shifting 27
Consider a time-domain function,
𝑔𝑔 𝑡𝑡 To Laplace transform we’ve assumed = 0 for < 0, or equivalently𝑔𝑔 𝑡𝑡 multiplied by 1(𝑔𝑔) 𝑡𝑡 𝑡𝑡 To shift by an𝑡𝑡 amount, , in time, we must also multiply𝑔𝑔 by𝑡𝑡 a shifted step function,𝑎𝑎 1
K. Webb 𝑡𝑡 − 𝑎𝑎 ESE 330 Time Shifting – Laplace Transform 28
The transform of the shifted function is given by
1 = ∞ −𝑠𝑠𝑠𝑠 ℒ 𝑔𝑔 𝑡𝑡 − 𝑎𝑎 � 𝑡𝑡 − 𝑎𝑎 �𝑎𝑎 𝑔𝑔 𝑡𝑡 − 𝑎𝑎 𝑒𝑒 𝑑𝑑𝑑𝑑 Performing a change of variables, let = and =
The transform becomes𝜏𝜏 𝑡𝑡 − 𝑎𝑎 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
1 = = = ∞ ∞ ∞ −𝑠𝑠 𝜏𝜏+𝑎𝑎 −𝑎𝑎𝑎𝑎 −𝑠𝑠𝜏𝜏 −𝑎𝑎𝑎𝑎 −𝑠𝑠𝑠𝑠 ℒ 𝑔𝑔 𝜏𝜏 � 𝜏𝜏 � 𝑔𝑔 𝜏𝜏 𝑒𝑒 𝑑𝑑𝜏𝜏 � 𝑔𝑔 𝜏𝜏 𝑒𝑒 𝑒𝑒 𝑑𝑑𝑑𝑑 𝑒𝑒 � 𝑔𝑔 𝜏𝜏 𝑒𝑒 𝑑𝑑𝑑𝑑 A shift by in the0 time domain corresponds0 to multiplication0 by in the Laplace domain −𝑎𝑎𝑎𝑎 𝑎𝑎 𝑒𝑒 1 = (13) −𝑎𝑎𝑎𝑎 ℒ 𝑔𝑔 𝑡𝑡 − 𝑎𝑎 � 𝑡𝑡 − 𝑎𝑎 𝑒𝑒 𝐺𝐺 𝑠𝑠 K. Webb ESE 330 Multiplication by time, 29 The Laplace transform of a function𝑡𝑡 multiplied by time: = ( ) (14) 𝑑𝑑 Consider a unitℒ 𝑡𝑡 ramp⋅ 𝑓𝑓 𝑡𝑡 function:− 𝑑𝑑𝑑𝑑 𝐹𝐹 𝑠𝑠 1 1 = 1 = = 𝑑𝑑 ℒ 𝑡𝑡 ℒ 𝑡𝑡 ⋅ 𝑡𝑡 − 2 Or a parabola: 𝑑𝑑𝑑𝑑 𝑠𝑠 𝑠𝑠 = = = 2 𝑑𝑑 1 2 2 3 In general ℒ 𝑡𝑡 ℒ 𝑡𝑡 ⋅ 𝑡𝑡 − 𝑑𝑑𝑑𝑑 𝑠𝑠 𝑠𝑠 ! = 𝑚𝑚 𝑚𝑚 ℒ 𝑡𝑡 𝑚𝑚+1 K. Webb 𝑠𝑠 ESE 330 Initial and Final Value Theorems 30
Initial Value Theorem Can determine the initial value of a time-domain signal or function from its Laplace transform
0 = lim (15)
𝑔𝑔 𝑠𝑠→∞ 𝑠𝑠𝑠𝑠 𝑠𝑠 Final Value Theorem Can determine the steady-state value of a time-domain signal or function from its Laplace transform
= lim (16)
𝑔𝑔 ∞ 𝑠𝑠→0 𝑠𝑠𝑠𝑠 𝑠𝑠 K. Webb ESE 330 Convolution 31
Convolution of two functions or signals is given by
= 𝑡𝑡
Result is a function𝑔𝑔 𝑡𝑡 of∗ time𝑥𝑥 𝑡𝑡 �0 𝑔𝑔 𝜏𝜏 𝑥𝑥 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝑑𝑑 is flipped in time and shifted by Multiply the flipped/shifted signal and the other signal Integrate𝑥𝑥 𝜏𝜏 the result from = 0 … 𝑡𝑡 May seem like an odd, arbitrary function now, but we’ll later see why it is very important𝜏𝜏 𝑡𝑡 Convolution in the time domain corresponds to multiplication in the Laplace domain
= (17)
K. Webb ℒ 𝑔𝑔 𝑡𝑡 ∗ 𝑥𝑥 𝑡𝑡 𝐺𝐺 𝑠𝑠 𝑋𝑋 𝑠𝑠 ESE 330 Impulse Function 32
Another common way to describe a dynamic system is with its impulse response System output in response to an impulse function input Impulse function defined by = 0, 0
𝛿𝛿 𝑡𝑡 = 1𝑡𝑡 ≠ ∞ � 𝛿𝛿 𝑡𝑡 𝑑𝑑𝑑𝑑 An infinitely−∞ tall, infinitely narrow pulse
K. Webb ESE 330 Impulse Function – Laplace Transform 33
To derive , consider the following function 1 , 0 ℒ= 𝛿𝛿 𝑡𝑡 0, < 0 or≤ 𝑡𝑡 ≤> 𝑡𝑡0 𝑔𝑔 𝑡𝑡 �𝑡𝑡0 𝑡𝑡 𝑡𝑡 𝑡𝑡0 Can think of as the sum of two step functions: 1 1 𝑔𝑔 𝑡𝑡 = 1 1 𝑔𝑔 𝑡𝑡 𝑡𝑡 − 𝑡𝑡 − 𝑡𝑡0 The transform of the first term𝑡𝑡0 is 𝑡𝑡0 1 1 1 =
ℒ 𝑡𝑡 Using the time-shifting property,𝑡𝑡0 the second𝑡𝑡0𝑠𝑠 term transforms to 1 1 = −𝑡𝑡0𝑠𝑠 0 𝑒𝑒 ℒ − 0 𝑡𝑡 − 𝑡𝑡 − 0 K. Webb 𝑡𝑡 𝑡𝑡 𝑠𝑠 ESE 330 Impulse Function – Laplace Transform 34
In the limit, as 0, , so
𝑡𝑡0 → 𝑔𝑔 𝑡𝑡 →=𝛿𝛿 lim𝑡𝑡
ℒ 𝛿𝛿 𝑡𝑡 𝑡𝑡0→01ℒ 𝑔𝑔 𝑡𝑡 = lim −𝑡𝑡0𝑠𝑠 − 𝑒𝑒 ℒ 𝛿𝛿 𝑡𝑡 𝑡𝑡0→0 Apply l’Hôpital’s rule 𝑡𝑡0𝑠𝑠 1 = lim −𝑡𝑡0𝑠𝑠 = lim = 𝑑𝑑 −𝑡𝑡0𝑠𝑠 − 𝑒𝑒 𝑑𝑑𝑡𝑡0 𝑠𝑠𝑒𝑒 𝑠𝑠 ℒ 𝛿𝛿 𝑡𝑡 𝑡𝑡0→0 𝑡𝑡0→0 𝑑𝑑 𝑡𝑡0𝑠𝑠 𝑠𝑠 𝑠𝑠 The Laplace transform of an𝑑𝑑 impulse𝑡𝑡0 function is one
= 1 (18)
K. Webb ℒ 𝛿𝛿 𝑡𝑡 ESE 330 Common Laplace Transforms 35
𝒈𝒈 𝒕𝒕 𝑮𝑮1𝒔𝒔 𝒈𝒈sin𝒕𝒕 𝑮𝑮 𝒔𝒔 + + −𝑎𝑎𝑎𝑎 𝜔𝜔 𝛿𝛿 𝑡𝑡 1 𝑒𝑒 𝜔𝜔𝜔𝜔 +2 2 1 cos 𝑠𝑠 𝑎𝑎 𝜔𝜔 + + −𝑎𝑎𝑎𝑎 𝑠𝑠 𝑎𝑎 𝑡𝑡 1 𝑒𝑒 𝜔𝜔𝜔𝜔 2 2 𝑠𝑠 ( ) 𝑠𝑠 𝑎𝑎 (𝜔𝜔0)
𝑡𝑡 2! 𝑔𝑔̇ 𝑡𝑡 𝑠𝑠𝑠𝑠 𝑠𝑠 − 𝑔𝑔 𝑠𝑠 0 0 𝑚𝑚 𝑚𝑚 2 𝑡𝑡 𝑚𝑚1+1 𝑔𝑔̈ 𝑡𝑡 𝑠𝑠 𝐺𝐺 𝑠𝑠 −1𝑠𝑠𝑠𝑠 − 𝑔𝑔̇ 𝑠𝑠 + 𝑡𝑡 −𝑎𝑎𝑎𝑎 𝑒𝑒 1 � 𝑔𝑔 𝜏𝜏 𝑑𝑑𝑑𝑑 𝐺𝐺 𝑠𝑠 𝑠𝑠 𝑎𝑎 0 𝑠𝑠 + + −𝑎𝑎𝑎𝑎 −𝑎𝑎𝑎𝑎 𝑡𝑡𝑒𝑒 2 𝑒𝑒 𝑔𝑔 𝑡𝑡 𝐺𝐺 𝑠𝑠 𝑎𝑎 sin 𝑠𝑠 𝑎𝑎 1 + 𝜔𝜔 −𝑎𝑎𝑎𝑎 2 2 cos 𝜔𝜔𝜔𝜔 𝑔𝑔 𝑡𝑡 − 𝑎𝑎 � 𝑡𝑡 − 𝑎𝑎 𝑒𝑒 𝐺𝐺 𝑠𝑠 𝑠𝑠 + 𝜔𝜔 𝑠𝑠 𝑑𝑑 2 2 K. Webb 𝜔𝜔𝜔𝜔 𝑡𝑡 � 𝑔𝑔 𝑡𝑡 − 𝐺𝐺 𝑠𝑠 ESE 330 𝑠𝑠 𝜔𝜔 𝑑𝑑𝑑𝑑 Example – Piecewise Function Laplace Transform
36
Determine the Laplace transform of a piecewise function:
A summation of functions with known transforms: Ramp Pulse – sum of positive and negative steps Transform is the sum of the individual, known transforms
K. Webb ESE 330 Example – Piecewise Function Laplace Transform
37 Treat the piecewise function as a sum of individual functions = +
𝑓𝑓 𝑡𝑡 𝑓𝑓1 𝑡𝑡 𝑓𝑓2 𝑡𝑡 Time-shifted, gated ramp 𝑓𝑓1 𝑡𝑡 Time-shifted pulse 2 𝑓𝑓 𝑡𝑡Sum of staggered positive and negative steps
K. Webb ESE 330 Example – Piecewise Function Laplace Transform
38
: time-shifted, gated ramp Ramp w/ slope of 2: 𝑓𝑓1 𝑡𝑡 = 2
Time-shifted𝑟𝑟 𝑡𝑡 ramp:⋅ 𝑡𝑡 = 2 ( 1)
Gating𝑟𝑟 𝑠𝑠function𝑡𝑡 ⋅ 𝑡𝑡 − Unity-amplitude pulse: = 1 1 1 2
Gate the𝑔𝑔 𝑡𝑡shifted𝑡𝑡 ramp:− − 𝑡𝑡 − =
𝑓𝑓1 𝑡𝑡 = 𝑟𝑟2𝑠𝑠 𝑡𝑡 ⋅ 𝑔𝑔1𝑡𝑡 1 1 1 2 1 K. Webb𝑓𝑓 𝑡𝑡 ⋅ 𝑡𝑡 − ⋅ 𝑡𝑡 − − 𝑡𝑡 − ESE 330 Example – Piecewise Function Laplace Transform
39
: time-shifted pulse Sum of staggered positive and 2 𝑓𝑓 negative𝑡𝑡 steps Positive step delayed by 2 sec: = 2 1( 2)
Negative𝑠𝑠2 𝑡𝑡step delayed⋅ 𝑡𝑡 − by 3 sec: = 2 1( 3)
Time-shifted𝑠𝑠3 𝑡𝑡 pulse− ⋅ 𝑡𝑡 − = + = 2 1 2 2 1( 3) 𝑓𝑓2 𝑡𝑡 𝑠𝑠2 𝑡𝑡 𝑠𝑠3 𝑡𝑡 K. Webb𝑓𝑓2 𝑡𝑡 ⋅ 𝑡𝑡 − − ⋅ 𝑡𝑡 − ESE 330 Example – Piecewise Function Laplace Transform
40
Sum the two individual time-domain functions = + = 2 1 1 1 1 2 + 2 1 2 2 1( 3) 𝑓𝑓 𝑡𝑡 𝑓𝑓1 𝑡𝑡 𝑓𝑓2 𝑡𝑡 = 2 1 1 1 𝑓𝑓 𝑡𝑡 ⋅ 𝑡𝑡 − ⋅ 𝑡𝑡 − − 𝑡𝑡 − ⋅ 𝑡𝑡 − − ⋅ 𝑡𝑡 − 2 1 2 𝑓𝑓 𝑡𝑡 𝑡𝑡 − ⋅ 𝑡𝑡 − +4 1 2 − 𝑡𝑡 ⋅ 𝑡𝑡 − +2 1 3 𝑡𝑡 − Transform the 𝑡𝑡individual− terms in = 2 1 1 1 + 2 1 2 𝑓𝑓 𝑡𝑡 𝐹𝐹 𝑠𝑠 ℒ 𝑡𝑡 − ⋅ 𝑡𝑡 − + +4 1 2 ℒ − 𝑡𝑡 ⋅ 𝑡𝑡 − + +2 1 3 ℒ 𝑡𝑡 − K. Webb ℒ 𝑡𝑡 − ESE 330 Example – Piecewise Function Laplace Transform
41
First term is a time-shifted ramp function 2 2 1 1 1 = −𝑠𝑠 𝑒𝑒 ℒ 𝑡𝑡 − ⋅ 𝑡𝑡 − 2 The next term is a time-shifted 𝑠𝑠step function multiplied by time
2 1 2 = 2 −2𝑠𝑠 𝑑𝑑 𝑒𝑒 ℒ − 𝑡𝑡 ⋅ 𝑡𝑡 − 𝑑𝑑𝑑𝑑 𝑠𝑠 2 = 2 + −2𝑠𝑠 −2𝑠𝑠 𝑒𝑒 𝑒𝑒 − 2 𝑠𝑠 𝑠𝑠 K. Webb ESE 330 Example – Piecewise Function Laplace Transform
42
The last two terms are time-shifted step functions 4 2 4 1 2 + 2 1 3 = + −2𝑠𝑠 −3𝑠𝑠 𝑒𝑒 𝑒𝑒 ℒ ⋅ 𝑡𝑡 − ⋅ 𝑡𝑡 − The piecewise function in the Laplace𝑠𝑠 domain:𝑠𝑠 2 2 4 2 = 2 + + + −𝑠𝑠 −2𝑠𝑠 −2𝑠𝑠 −2𝑠𝑠 −3𝑠𝑠 𝑒𝑒 𝑒𝑒 𝑒𝑒 𝑒𝑒 𝑒𝑒 𝐹𝐹 𝑠𝑠 2 − 2 2𝑠𝑠 2 𝑠𝑠 2𝑠𝑠 𝑠𝑠 𝑠𝑠 = + −𝑠𝑠 −2𝑠𝑠 −3𝑠𝑠 𝑒𝑒 𝑒𝑒 𝑒𝑒 𝐹𝐹 𝑠𝑠 2 − 2 𝑠𝑠 𝑠𝑠 𝑠𝑠
K. Webb ESE 330 43 Inverse Laplace Transform
We’ve just seen how time-domain functions can be transformed to the Laplace domain. Next, we’ll look at how we can solve differential equations in the Laplace domain and transform back to the time domain.
K. Webb ESE 330 Laplace Transforms – Differential Equations
44 Consider the simple spring/mass/damper system from the previous section of notes State equations are:
= + (1) 𝑏𝑏 𝑝𝑝̇ = − 𝑚𝑚 𝑝𝑝 − 𝑘𝑘𝑘𝑘 𝐹𝐹𝑖𝑖𝑖𝑖 𝑡𝑡 (2) 1 Taking the displacement𝑥𝑥̇ 𝑚𝑚 𝑝𝑝 of the mass as the output = (3)
Using (2) and 𝑦𝑦(3) in𝑥𝑥 (1) we get a single second-order differential equation + + = (4) 𝑏𝑏 𝑘𝑘 1 K. Webb 𝑦𝑦̈ 𝑚𝑚 𝑦𝑦̇ 𝑚𝑚 𝑦𝑦 𝑚𝑚 𝐹𝐹𝑖𝑖𝑖𝑖 𝑡𝑡 ESE 330 Laplace Transforms – Differential Equations
45
We’ll now use Laplace transforms to determine the step response of the system 1N step force input , < 0 = 1 = , 0 (5) 0𝑁𝑁 𝑡𝑡 𝐹𝐹𝑖𝑖𝑖𝑖 𝑡𝑡 1𝑁𝑁 � 𝑡𝑡 � For the step response,1𝑁𝑁 𝑡𝑡 ≥we assume zero initial conditions
0 = 0 and 0 = 0 (6)
Using the derivative𝑦𝑦 property𝑦𝑦̇ of the Laplace transform, (4) becomes 1 0 0 + (0) + = 2 𝑏𝑏 𝑏𝑏 𝑘𝑘 𝑠𝑠 𝑌𝑌 𝑠𝑠 − 𝑠𝑠𝑠𝑠 − 𝑦𝑦̇ 𝑠𝑠𝑌𝑌 𝑠𝑠 − 𝑦𝑦 𝑌𝑌 𝑠𝑠 𝐹𝐹𝑖𝑖𝑖𝑖 𝑠𝑠 + + 𝑚𝑚= 𝑚𝑚 𝑚𝑚 𝑚𝑚 (7) 2 𝑏𝑏 𝑘𝑘 1 𝑖𝑖𝑖𝑖 K. Webb 𝑠𝑠 𝑌𝑌 𝑠𝑠 𝑚𝑚 𝑠𝑠𝑌𝑌 𝑠𝑠 𝑚𝑚 𝑌𝑌 𝑠𝑠 𝑚𝑚 𝐹𝐹 𝑠𝑠 ESE 330 Laplace Transforms – Differential Equations
46
The input is a step, so (7) becomes
+ + = (8) 2 𝑏𝑏 𝑘𝑘 1 1 𝑠𝑠 𝑌𝑌 𝑠𝑠 𝑚𝑚 𝑠𝑠𝑌𝑌 𝑠𝑠 𝑚𝑚 𝑌𝑌 𝑠𝑠 𝑚𝑚 1𝑁𝑁 𝑠𝑠 Solving (8) for
1 1 𝑌𝑌 𝑠𝑠 + + = 2 𝑏𝑏 𝑘𝑘 𝑌𝑌 𝑠𝑠 𝑠𝑠 𝑠𝑠 / = 𝑚𝑚 𝑚𝑚 𝑚𝑚 𝑠𝑠 (9) 1 𝑚𝑚 2 𝑏𝑏 𝑘𝑘 Equation (9) is the𝑌𝑌 𝑠𝑠solution𝑠𝑠 𝑠𝑠 to+𝑚𝑚 the𝑠𝑠+𝑚𝑚 differential equation of (4), given the step input and I.C.’s The system step response in the Laplace domain Next, we need to transform back to the time domain
K. Webb ESE 330 Laplace Transforms – Differential Equations
47 / = (9) 1 𝑚𝑚 2 𝑏𝑏 𝑘𝑘 𝑌𝑌 𝑠𝑠 𝑠𝑠 𝑠𝑠 + 𝑠𝑠+ The form of (9) is typical𝑚𝑚 of𝑚𝑚 Laplace transforms when dealing with linear systems
A rational polynomial in Here, the numerator is 0th-order 𝑠𝑠 = 𝐵𝐵 𝑠𝑠 𝑌𝑌 𝑠𝑠 Roots of the numerator polynomial,𝐴𝐴 𝑠𝑠 , are called the zeros of the function Roots of the denominator polynomial,𝐵𝐵 𝑠𝑠 , are called the poles of the function
K. Webb 𝐴𝐴 𝑠𝑠 ESE 330 Inverse Laplace Transforms 48 / = (9) 1 𝑚𝑚 2 𝑏𝑏 𝑘𝑘 𝑌𝑌 𝑠𝑠 𝑠𝑠 𝑠𝑠 + 𝑠𝑠+ To get (9) back into the𝑚𝑚 time𝑚𝑚 domain, we need to perform an inverse Laplace transform An integral inverse transform exists, but we don’t use it Instead, we use partial fraction expansion
Partial fraction expansion Idea is to express the Laplace transform solution, (9), as a sum of Laplace transform terms that appear in the table Procedure depends on the type of roots of the denominator polynomial Real and distinct Repeated Complex
K. Webb ESE 330 Inverse Laplace Transforms – Example 1
49
Consider the following system parameters =
𝑚𝑚= 1𝑘𝑘𝑘𝑘 16𝑁𝑁 𝑘𝑘 = 10𝑚𝑚 𝑁𝑁 � 𝑠𝑠 𝑏𝑏 Laplace transform𝑚𝑚 of the step response becomes
= (10) 1 2 𝑠𝑠 𝑠𝑠 +10𝑠𝑠+16 Factoring the denominator𝑌𝑌 𝑠𝑠
= (11) 1 𝑠𝑠 𝑠𝑠+2 𝑠𝑠+8 In this case, the denominator𝑌𝑌 𝑠𝑠 polynomial has three real, distinct roots = 0, = 2, = 8
K. Webb ESE 330 𝑠𝑠1 𝑠𝑠2 − 𝑠𝑠3 − Inverse Laplace Transforms – Example 1
50
Partial fraction expansion of (11) has the form = = + + (12) 1 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 𝑌𝑌 𝑠𝑠The numerator𝑠𝑠 𝑠𝑠+2 𝑠𝑠+ 8coefficients,𝑠𝑠 𝑠𝑠+ 2 , 𝑠𝑠,+ and8 , are called residues 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 Can already see the form of the time-domain function Sum of a constant and two decaying exponentials
To determine the residues, multiply both sides of (12) by the denominator of the left-hand side 1 = + 2 + 8 + + 8 + + 2
1 = 𝑟𝑟1 𝑠𝑠 + 10 𝑠𝑠 + 16 𝑟𝑟2+𝑠𝑠 𝑠𝑠 + 8 𝑟𝑟3𝑠𝑠+𝑠𝑠 + 2 2 2 2 Collecting terms,𝑟𝑟1𝑠𝑠 we 𝑟𝑟have1𝑠𝑠 𝑟𝑟1 𝑟𝑟2𝑠𝑠 𝑟𝑟2𝑠𝑠 𝑟𝑟3𝑠𝑠 𝑟𝑟3𝑠𝑠 1 = + + + 10 + 8 + 2 + 16 (13) 2 K. Webb 𝑠𝑠 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 𝑠𝑠 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 𝑟𝑟1 ESE 330 Inverse Laplace Transforms – Example 1
51
Equating coefficients of powers of on both sides of (13) gives a system of three equations in three unknowns 𝑠𝑠 : + + = 0 2: 10 + 8 + 2 = 0 𝑠𝑠 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 1: 16 = 1 𝑠𝑠 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 0 Solving for𝑠𝑠 the residues𝑟𝑟1 gives = 0.0625 = 0.0833 𝑟𝑟1 = 0.0208 𝑟𝑟2 − The Laplace transform𝑟𝑟3 of the step response is . . . = + (14) 0 0625 0 0833 0 0208 Equation (14) can𝑌𝑌 now𝑠𝑠 be transformed𝑠𝑠 − 𝑠𝑠+2 back𝑠𝑠 +to8 the time domain using the Laplace transform table
K. Webb ESE 330 Inverse Laplace Transforms – Example 1
52
The time-domain step response of the system is the sum of a constant term and two decaying exponentials:
= 0.0625 0.0833 + 0.0208 (15) −2𝑡𝑡 −8𝑡𝑡 𝑦𝑦 𝑡𝑡 − 𝑒𝑒 𝑒𝑒
Step response plotted in MATLAB
Characteristic of a signal having only real poles No overshoot/ringing
Steady-state displacement agrees with intuition force applied to a / spring 1𝑁𝑁 16𝑁𝑁 𝑚𝑚
K. Webb ESE 330 Inverse Laplace Transforms – Example 1
53
Go back to (10) and apply the initial value theorem 1 0 = lim = lim = + + 16
𝑦𝑦 𝑠𝑠→∞ 𝑠𝑠𝑌𝑌 𝑠𝑠 𝑠𝑠→∞ 2 0𝑐𝑐𝑐𝑐 Which is, in fact our assumed𝑠𝑠 10𝑠𝑠 initial condition
Next, apply the final value theorem to the Laplace transform step response, (10) 1 = lim = lim + + 16
𝑦𝑦 ∞ 𝑠𝑠→0 𝑠𝑠𝑌𝑌 𝑠𝑠 𝑠𝑠→0 2 = = 0. = 𝑠𝑠6. 10𝑠𝑠 1 𝑦𝑦 ∞ 16 0625𝑚𝑚 25𝑐𝑐𝑐𝑐 This final value agrees with both intuition and our numerical analysis
K. Webb ESE 330 Inverse Laplace Transforms – Example 2
54
Reduce the damping and re-calculate the step response =
𝑚𝑚= 1𝑘𝑘𝑘𝑘 16𝑁𝑁 𝑘𝑘 = 8𝑚𝑚 𝑁𝑁 � 𝑠𝑠 𝑏𝑏 Laplace transform𝑚𝑚 of the step response becomes = (16) 1 2 Factoring the denominator𝑌𝑌 𝑠𝑠 𝑠𝑠 𝑠𝑠 +8𝑠𝑠+16 = (17) 1 2 In this case, the denominator𝑌𝑌 𝑠𝑠 𝑠𝑠 𝑠𝑠+ polynomial4 has three real roots, two of which are identical = 0, = 4, = 4
K. Webb 𝑠𝑠1 𝑠𝑠2 − 𝑠𝑠3 − ESE 330 Inverse Laplace Transforms – Example 2
55
Partial fraction expansion of (17) has the form
= = + + (18) 1 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 2 2 𝑌𝑌 𝑠𝑠 𝑠𝑠 𝑠𝑠+4 𝑠𝑠 𝑠𝑠+4 𝑠𝑠+4 Again, find residues by multiplying both sides of (18) by the left- hand side denominator 1 = + 4 + + 4 + 2 1 = 𝑟𝑟1 𝑠𝑠 + 8 +𝑟𝑟216𝑠𝑠 𝑠𝑠 + 𝑟𝑟+3𝑠𝑠4 + 2 2 Collecting𝑟𝑟 terms,1𝑠𝑠 we𝑟𝑟1 𝑠𝑠have 𝑟𝑟1 𝑟𝑟2𝑠𝑠 𝑟𝑟2𝑠𝑠 𝑟𝑟3𝑠𝑠 1 = + + 8 + 4 + + 16 (19) 2 1 2 1 2 3 1 K. Webb 𝑠𝑠 𝑟𝑟 𝑟𝑟 𝑠𝑠 𝑟𝑟 𝑟𝑟 𝑟𝑟 𝑟𝑟 ESE 330 Inverse Laplace Transforms – Example 2
56
Equating coefficients of powers of on both sides of (19) gives a system of three equations in three unknowns 𝑠𝑠 : + = 0 2: 8 + 4 + = 0 𝑠𝑠 𝑟𝑟1 𝑟𝑟2 1: 16 = 1 𝑠𝑠 𝑟𝑟1 𝑟𝑟2 𝑟𝑟3 0 Solving for the residues gives 𝑠𝑠 𝑟𝑟1 = 0.0625 = 0.0625 𝑟𝑟1 = 0.2500 𝑟𝑟2 − The Laplace transform𝑟𝑟3 of− the step response is . . . = (20) 0 0625 0 0625 0 25 2 Equation (20) can 𝑌𝑌now𝑠𝑠 be transformed𝑠𝑠 − 𝑠𝑠+4 back− 𝑠𝑠 to+4 the time domain using the Laplace transform table
K. Webb ESE 330 Inverse Laplace Transforms – Example 2
57
The time-domain step response of the system is the sum of a constant, a decaying exponential, and a decaying exponential scaled by time:
= 0.0625 0.0625 0. (21) −4𝑡𝑡 −4𝑡𝑡 𝑦𝑦 𝑡𝑡 − 𝑒𝑒 − 25𝑡𝑡𝑒𝑒
Step response plotted in MATLAB
Again, characteristic of a signal having only real poles
Similar to the last case
A bit faster – slow pole at = 2 was eliminated 𝑠𝑠 −
K. Webb ESE 330 Inverse Laplace Transforms – Example 3
58
Reduce the damping even further and go through the process once again =
𝑚𝑚= 1𝑘𝑘𝑘𝑘 16𝑁𝑁 𝑘𝑘 = 4𝑚𝑚 𝑁𝑁 � 𝑠𝑠 𝑏𝑏 Laplace transform𝑚𝑚 of the step response becomes = (22) 1 2 The second-order term𝑌𝑌 𝑠𝑠 in the𝑠𝑠 𝑠𝑠 +denominator4𝑠𝑠+16 now has complex roots, so we won’t factor any further The denominator polynomial still has a root at zero and now has two roots which are a complex-conjugate pair = 0, = 2 + .464, = 2 .464
𝑠𝑠1 𝑠𝑠2 − 𝑗𝑗푗 𝑠𝑠3 − − 𝑗𝑗푗 K. Webb ESE 330 Inverse Laplace Transforms – Example 3
59
Want to cast the partial fraction terms into forms that appear in the Laplace transform table Second-order terms should be of the form
( ) (23) 𝑟𝑟𝑖𝑖 𝑠𝑠+𝜎𝜎 +𝑟𝑟𝑖𝑖+1𝜔𝜔 2 2 𝑠𝑠+𝜎𝜎 +𝜔𝜔 This will transform into the sum of damped sine and cosine terms + + = cos + sin + + + + −1 𝑠𝑠 𝜎𝜎 𝜔𝜔 −𝜎𝜎𝜎𝜎 −𝜎𝜎𝜎𝜎 ℒ 𝑟𝑟𝑖𝑖 2 2 𝑟𝑟𝑖𝑖+1 2 2 𝑟𝑟𝑖𝑖𝑒𝑒 𝜔𝜔𝜔𝜔 𝑟𝑟𝑖𝑖+1𝑒𝑒 𝜔𝜔𝜔𝜔 𝑠𝑠 𝜎𝜎 𝜔𝜔 𝑠𝑠 𝜎𝜎 𝜔𝜔 To get the second-order term in the denominator of (22) into the form of (23), complete the square, to give the following partial fraction expansion
. = = + . (24) 1 𝑟𝑟1 𝑟𝑟2 𝑠𝑠+2 +𝑟𝑟3 3 464 2 2 2 𝑌𝑌 𝑠𝑠 𝑠𝑠 𝑠𝑠 +4𝑠𝑠+16 𝑠𝑠 𝑠𝑠+2 + 3 464 K. Webb ESE 330 Inverse Laplace Transforms – Example 3
60
Note that the and terms in (23) and (24) are the real and imaginary parts of the complex-conjugate𝜎𝜎 𝜔𝜔 denominator roots
, = ± = 2 ± .464
𝑠𝑠2 3 −𝜎𝜎 𝑗𝑗𝑗𝑗 − 𝑗𝑗푗 Multiplying both sides of (24) by the left-hand-side denominator, equate coefficients and solve for residues as before: = 0.0625 = 0.0625 𝑟𝑟1 = 0.0361 𝑟𝑟2 − 3 Laplace transform𝑟𝑟 of the− step response is
. . . . = (25) . . 0 0625 0 0625 𝑠𝑠+2 0 0361 3 464 2 2 2 2 𝑠𝑠 𝑠𝑠+2 + 3 464 𝑠𝑠+2 + 3 464 K. Webb 𝑌𝑌 𝑠𝑠 − − ESE 330 Inverse Laplace Transforms – Example 3
61
The time-domain step response of the system is the sum of a constant and two decaying sinusoids:
= 0.0625 0.0625 cos 3. 0.0361 sin(3. ) (26) −2𝑡𝑡 −2𝑡𝑡 𝑦𝑦 𝑡𝑡 − 𝑒𝑒 464𝑡𝑡 − 𝑒𝑒 464𝑡𝑡 Step response and individual components plotted in MATLAB
Characteristic of a signal having complex poles
Sinusoidal terms result in overshoot and (possibly) ringing
K. Webb ESE 330 Laplace-Domain Signals with Complex Poles
62
The Laplace transform of the step response in the last example had complex poles A complex-conjugate pair: = ±
Results in sine and cosine terms𝑠𝑠 − 𝜎𝜎in 𝑗𝑗𝑗𝑗 the time domain cos + sin −𝜎𝜎𝜎𝜎 −𝜎𝜎𝜎𝜎 Imaginary𝐴𝐴𝑒𝑒 part𝜔𝜔𝜔𝜔 of the𝐵𝐵𝑒𝑒 roots, 𝜔𝜔𝜔𝜔 Frequency of oscillation of sinusoidal components of the signal 𝜔𝜔
Real part of the roots, , Rate of decay of the sinusoidal components 𝜎𝜎
Much more on this later
K. Webb ESE 330